CODE-F
AIPMT – 2015 TEST PAPER WITH SOLUTIONS
(HELD ON SUNDAY 03th MAY, 2015)
1.
The reaction of C6H5CH = CHCH 3 with HBr
produces:(1) C6H5CH 2CHCH3
(2) C6H5CH2CH2CH2Br
Br
CH=CHCH3
(3)
(4)
C6H 5CHCH2CH3
3.
The Ksp of Ag2CrO 4, AgCl, AgBr and AgI are
respectively, 1.1 × 10–12, 1.8 × 10–10, 5.0 × 10–13,
8.3 × 10–17. Which one of the following salts will
precipitate last if AgNO3 solution is added to the
solution containing equal moles of NaCl, NaBr, NaI
and Na2CrO4 ?
(1) AgCl
(2) AgBr
(3) Ag2CrO4 (4) AgI
Ans. (3)
Sol.
Br
Br
Þ ëé Ag+ ûù =
Ans. (4)
éëCrO ùû
-2
4
=
1.1 ´ 10-12
m
éëCrO4-2 ùû = max
CH CH
Å
CH
EN
C6H5
4.
Bithional is generally added to the soaps as an
additive to function as a/an :(1) Dryer
(2) Buffering agent
(3) Antiseptic
(4) Softner
Ans. (3)
Sol. Bithionol (the compound is also called bithional) is
added to soaps to impart antiseptic properties
CH3+HBr
CH2 CH3
Br
CH CH2 CH3
LL
C6H5
Br
A
In Duma's method for estimation of nitrogen, 0.25
g of an organic compound gave 40 mL of nitrogen
collected at 300 K temperature and 725 mm pressure.
If the aqueous tension at 300 K is 25 mm, the
percentage of nitrogen in the compound is :(1) 18.20 (2) 16.76 (3) 15.76 (4) 17.36
Ans. (2)
Sol. Volume of nitrogen collected at 300 K and 725 mm
pressure is 40 mL actual pressure
= 725 – 25 = 700 mm
273 × 700 × 40
= 33.52mL
=
300 ´ 760
22,400 mL of N2 at STP weight = 28 g
33.5 mL of nitrogen weight =
Percentage of nitrogen
28 ´ 33.52 ´ 100
= 16.76%
22400 ´ 0.25
5.
Cl
OH
HO
Cl
S
Cl
Cl
Bithionol
"Metals are usually not found as nitrates in their
ores".
Out of the following two (a and b) reasons which
is/are true for the above observation ?
(a) Metal nitrates are highly unstable.
Volume of nitrogen at STP
=
K sp
so answer is Ag2CrO4
Sol. C H
6 5
2.
2
Ag2 CrO 4 ® K sp = ëé Ag + ûù ëéCrO4-2 ûù
28 ´ 33.52
g
22400
(b) Metal nitrates are highly soluble in water.
(1) a and b are false
(2) a is false but b is true
(3) a is true but b is false
(4) a and b are true
Ans. (2)
Sol. Metal nitrates are stable like NaNO3 and KNO3 and
highly soluble in water.
1
AIPMT-2015
6.
The correct bond order in the following species is:+
(1) O2+
2 < O2 < O2
(2) O+2 < O2- < O2+
2
(3) O2- < O2+ < O2+
2
+
(4) O2+
2 < O2 < O2
Ans. (3)
Sol.
No. of e -
No. of e -
O2+
O2+2
17
15
14
The species Ar, K+ and Ca2+ contain the same
number of electrons. In which order do their radii
increase ?
(1) Ca2+ < Ar < K+
(2) Ca2+ < K+ < Ar
(4) Ar < K+ < Ca2+
An organic compound 'X' having molecular formula
C5H10O yields phenyl hydrazone and gives negative
response to the Iodoform test and Tollen's test. It
produces n-pentane on reduction. 'X' could be :(1) 2-pentanone
(2) 3-pentanone
(3) n-amyl alcohol
(4) pentanal
Ans. (2)
Ph—NH—NH2
Sol.
(1)
T
In K
vs.T
vs.
l
T
(2) In K vs.
l
Ans. (2)
Sol. Arrhenius equation
K = A.e - Ea/RT Þ lnK = lnA -
9.
Ea
RT
(3)
2
1
T
ClO3- ,SO23 -
(2) SO23 , NO3
(4)
n-pentane
According to questions CH3 CH2 C CH2 CH3
O
does not give iodoform as well as Tollen’s test
H3C CH2 C
CH2 CH3
O
reduction
CH3—CH2—CH2—CH2—CH3
Zn-Hg/HCl
n-pentane
—
or NH2-NH2/OH
Which of the following options represents the correct
bond order ?
(2) O -2 > O2 < O2+
(3) O -2 < O2 > O+2
(4) O -2 > O2 > O 2+
Which of the following pairs of ions are isoelectronic
and isostructural ?
(1) ClO -3 ,CO32 -
11.
Negative
(1) O -2 < O2 < O +2
so, activation energy of reaction can be determined
from the slope of lnK vs
Tollen's test
T
(4) In K vs. T
A
(3)
T
LL
The activation energy of a reaction can be
determined from the slope of which of the following
graphs ?
Phenyl hydrazone
I2 + NaOH
Negative
Iodoform test
Reduction
hence increasing order of radius is Ca+2 < K+ < Ar
In K
sp3
both are having one lone pair on central atom hence
they are pyramidal.
X compound
(C5H10O)
1
Atomic radius µ Z
eff
8.
42
3
EN
(3) K+ < Ar < Ca2+
Ans. (2)
Sol. In isoelectronic species
42
10.
Bond order 1.5 2.5 3.0
7.
SO3-2
hybridisation sp
Ans. (3)
Sol. According to molecular orbital theory (MOT)
O2-
ClO3-
CO23 - ,SO32 -
Ans. (1)
Sol. According to molecular orbital theory (MOT)
No. of e -
O2-
O2
17
16 15
Bond order 1.5 2
O2+
2.5
CODE-F
12.
O with
Treatment of cyclopentanone
Sol. FCC : r =
16.
methyl lithium gives which of the following species?
(1) Cyclopentanonyl cation
(2) Cyclopentanonyl radical
(3) Cyclopentanonyl biradical
(4) Cyclopentanonyl anion
Ans. (4)
CH3
Å
O + CH3Li
Sol.
13.
NH2
Sol.
electrolytic
nitrobenzene
strongly
NH2
rearrangement
Phenyl hydroxyl amine
OH
p-Aminophenol
H2
H
C = C – C– Cl
H
(3) H3C
H2
H
C = C – C– Cl
H
(4) H3C
H2
H
C = C – C– Cl
H
CH
CH
CH2 Cl
Which one of the following electrolytes has the same
value of van't Hoff's factor (i) as that of the Al2(SO4)3
(if all are 100% ionised) ?
(1) K3[Fe(CN)6]
(2) Al(NO3)3
(3) K4[Fe(CN)6]
(4) K2SO4
Ans. (3)
Sol. van’t Hoff factor of
A
Magnetic moment 2.84 B.M. is given by :(At. nos, Ni = 28, Ti = 22, Cr = 24, Co = 27)
(1) Ti3+
(2) Cr2+
(3) Co2+
(4) Ni2+
Ans. (4)
K 4 [ Fe ( CN )6 ] ® 4K + + [Fe ( CN )6 ]
hence m = 2.8 B.M., paramagnetic
A given metal crystallizes out with a cubic structure
having edge length of 361 pm. If there are four
metal atoms in one unit cell, what is the radius of
one atom?
(1) 127 pm
(2) 80 pm
(3) 108 pm
(4) 40 pm
Ans. (1)
-4
so n =5 Þ i = n =5
number of unpaired electrons (n) =2
Q m = n (n + 2)
Al 2 ( SO 4 )3 ® 2Al +3 + 3SO -42
so n = 5
Sol. Ni+2 = 3d8
15.
(2) H3C
17.
aniline
HN OH
acidic medium
14.
H2
H
C = C – C– Cl
H
Ans. (2)
Sol. After leaving Cl–, due to resonance, p bond is also
transferred
LL
reduction
(1) H3C
H3C
weakly acidic
medium
NO2
361
= 127pm
2 ´ 1.4141
EN
The electrolytic reduction of nitrobenzene in strongly
acidic medium produces :(1) Azoxybenzene
(2) Azobenzene
(3) Aniline
(4) p-Aminophenol
Ans. (4)
2 2
=
Which of the following is the most correct electron
displacement for a nucleophilic reaction to take
place?
O
cyclopentanonyl anion
a
18.
Nitrogen dioxide and sulphur dioxide have some
properties in common. Which property is shown by
one of these compounds, but not by the other ?
(1) is a reducing agent
(2) is soluble in water
(3) is used as a food-preservative
(4) forms 'acid-rain'
Ans. (3)
Sol. NO2 is not used as food preservative.
3
AIPMT-2015
19.
The total number of p-bond electrons in the following
structure is :H
H3C
H
H
CH3
H3C
H
H2C
(1) 8
Ans. (1)
Sol. H3C
(2) 12
H H
CH3
(3) 16
H
CH3
H3C
H2C
H
(4) 4
23.
The number of d-electrons in Fe2+ (Z = 26) is not
equal to the number of electrons in which one of
the following?
(1) p-electrons in Cl (Z = 17)
(2) d-electrons in Fe (Z = 26)
(3) p-electrons in Ne (Z = 10)
(4) s-electrons in Mg (Z = 12)
Ans. (1)
Sol. Fe+2 = 3d6 (number of ‘d’ electrons = 6)
in Cl = 1s2 2s2 2p6 2s2 3p5
Total p bonds = 4
Total p electrons = 8
24.
CH3
20.
Solubility of the alkaline earth's metal sulphates in
water decreases in the sequence :(1) Ca > Sr > Ba > Mg (2) Sr > Ca > Mg > Ba
(3) Ba > Mg > Sr > Ca (4) Mg > Ca > Sr > Ba
Ans. (4)
Sol. Due to very small size of Mg +2, Mg +2 shows
maximum hydration energy.
total p electrons = 11, which are not equal to number of ‘d’ electrons in Fe+2
In which of the following compounds, then C - Cl
bond ionisation shall give most stable carbonium ion?
H
H3C
Hydration
energy
Maximum bond angle at nitrogen is present in which
of the following ?
(1)
NO -2
(3) NO3-
(2)
NO +2
(4) NO 2
O2NH2C
Sol.
H3C
C Cl
A
¾¾¾¾¾¾¾
®
The value of K is very high so the system will contain mostly products at equilibrium.
4
H
C – Cl
H3C
H3C
C
+
H3C CH3
t-butyl carbonium ion
H
NO-2 < NO-3 < NO2 < NO 2+
bond angle­
If the value of an equilibrium constant for a particular
reaction is 1.6 × 1012, then at equilibrium the system
will contain :(1) mostly reactants
(2) mostly products
(3) similar amounts of reactants products
(4) all reactants
Ans. (2)
Sol. The value of equilibrium constant for reaction
K = 1.6 × 1012
(4)
Cl
H3C CH3
Ans. (2)
Sol. NO2+ : sp hybridisation (bond angle = 180°)
22.
H3C
C – Cl
H
Ans. (2)
LL
21.
(3)
Solubility
CH3
H
MgSO4 > CaSO4 > SrSO 4 > BaSO 4
CH – Cl
(2)
EN
(1) H3C
C – Cl
CH
Cl
H
Cl
CH
Å
benzyl carbonium ion
[stable due to resonance]
H
H
Cl
C Cl
O2NCH2 H
O2NH2C H
H3C
H3C
-I effect
H
C Cl
CÅ
Cl
H3C
H3C
H
CÅ
Secondary carbonium ion
[isopropyl carbonium ion]
CODE-F
Most stable carbonium ion is benzyl carbocation due
to resonance
Å
CH2
CH2
Å
Å
Å
CH2
CH2
25.
CH3
Ph
CH3 – C – CH –
Ph – C – Ph
CH3
(I)
(II)
– CH3
(III)
Sol.
C
.
CH
.
Ph
Ph
.
C
Ph
(I)
CH3
H
A
CH3
(III)
(II)
Only (III) has H in conjugation with free radical
27.
The reaction
CH3
CH3
CH3–C–ONa+CH3CH2Cl
CH3
AgNO3
® no ppt
ëéCo(NH3 )3 Cl3 ûù ¾¾¾¾
29. A mixture of gases contains H 2 and O2 gases in the
ratio of 1 : 4 (w/w). What is the molar ratio of the
two gases in the mixture ?
(1) 4 : 1 (2) 16 : 1 (3) 2 : 1
(4) 1 : 4
Ans. (1)
Sol.
CH3–C–O–CH2–CH3
–NaCl
w H2
w O2
=
nH
1
1/ 2
4
Þ 2 =
=
4
n O2 4 / 32 1
Which of the following processes does not involve
oxidation of iron ?
(1) Decolourization of blue CuSO4 solutution by iron
(2) Formation of Fe(CO)5 from Fe
(3) Liberation of H2 from steam by iron at high
temperature
(4) Rusting of iron sheets
Ans. (2)
LL
CH3
CoCl 3 .3NH 3 Þ éëCo ( NH 3 )3 Cl3 ùû
30.
Hyperconjugation occurs In :(1) II only (2) III only (3) I and III (4) I only
Ans. (2)
H3C
Sol.
EN
A device that converts energy of combustion of fuels
like hydrogen and methane, directly into electrical
energy is known as :(1) Electrolytic cell
(2) Dynamo
(3) Ni-Cd cell
(4) Fuel Cell
Ans. (4)
Sol. A device that converts energy of combustion of fuels,
directly into electrical energy is known as fuel cell.
26. Consider the following compounds
Sol. Given reaction is an important laboratory method
for the preparation of symmetrical and
unsymmetrical ethers. In this method, an alkyl halide
is allowed to react with sodium alkoxide.
28. Cobalt (III) chloride forms several octahedral
complexes with ammonia. Which of the following
will not give test of chloride ions with silver nitrate
at 25ºC ?
(1) CoCl3·4NH3
(2) CoCl3·5NH3
(3) CoCl3·6NH3
(4) CoCl3·3NH3
Ans. (4)
CH3
is called :(1) Williamson continuous etherification process
(2) Etard reaction
(3) Gatterman - Koch reaction
(4) Williamson Synthesis
Ans. (4)
Sol.
Fe0 + 5CO ® [ Fe0 ( CO )5 ]
No change in the oxidation state of iron
Because of lanthanoid contraction, which of the
following pairs of elements have nearly same atomic
radii ? (Nubmers in the parenthesis are atomic
numbers).
(1) Zr (40) and Nb (41) (2) Zr (40) and Hf (72)
(3) Zr (40) and Ta (73) (4) Ti (22) and Zr (40)
Ans. (2)
Sol. Due to lanthanoid contraction
atomic radii of Zr and Hf is almost similar.
32. Which of the following statements is correct for a
reversible process in a state of equilibrium ?
(1) DG = 2.30 RT log K
(2) DGº = –2.30 RT log K
(3) DGº = 2.30 RT log K
(4) DG = –2.30 RT log K
5
31.
AIPMT-2015
Ans. (2)
Sol.
DG0 = -2.30RT log K
because at equilibrium DG =0
33.
The angular momentum of electron in 'd' orbital is
equal to :-
2h
(1)
(2) 2 3 h
(3) 0 h
R
CH CH R + H2 (1 mole)
R
CH2 CH2 R + DH
Sol.
6h
(4)
The enthalpy of the hydrogenation of these
compounds will be in the order as :(1) III > II > I
(2) II > III > I
(3) II > I > III
(4) I > II > III
Ans. (1)
Ans. (4)
Sol. Orbital angular momentum =
for d-orbital l=2
l ( l + 1).h
so orbital angular momentum = 2 ( 2 + 1)h = 6h
34.
37.
1
stability of alkene
The enolic form of ethyl acetoacetate as below has:-
H3 C
H
C
C
C
O
H 3C
EN
The boiling point of 0.2 mol kg–1 solution of X in
water is greater than equimolal solution of Y in water.
Which one of the following statements is true in this
case ?
(1) Molecular mass of X is greater than the
molecular mass of Y.
(2) Molecular mass of X is less than the molecular
mass of Y.
(3) Y is undergoing dissociation in water while X
undergoes no change.
(4) X is undergoing dissociation is water.
Ans. (4)
enthalpy of hydrogenation µ
OH
same solvent so, Kb is same
m is same (given)
A
38.
so, x is undergoing dissociation in water.
35.
The function of "Sodium pump" is a biological
process operating in each and every cell of all
animals. Which of the following biologically
important ions is also a consituent of this pump :(1) Mg2+
(2) K+
(3) Fe2+
(4) Ca2+
Ans. (2)
Sol. K+ ion is a constituent of sodium pump
36. Given :CH
H 3C
CH
(I)
6
3
3
CH
H 3C
CH
(II)
3
2
CH
H 2C
CH
(III)
2
2
O
OC2H5
H
Sol. H
ix × k b × m > i y × k b × m Þ i x > i y
H2
C
C
O
OC2H5
(1) 16 sigma bonds and 1 pi - bond
(2) 9 sigma bonds and 2 pi - bonds
(3) 9 sigma bonds and 1 pi - bond
(4) 18 sigma bonds and 2 pi - bonds
Ans. (4)
LL
( DTb )x > ( DTb )y
Sol.
C
H
H H
C C C C O C C H
H O H O
H H
18 s + 2p
Biodegradable polymer which can be produced from
glycine and aminocaproic acid is :(1) PHBV
(2) Buna - N
(3) Nylon 6, 6
(4) Nylon 2- nylon 6
Ans. (4)
Sol. Nylon 2-Nylon-6 is an alternating polyamide
copolymer of glycine (NH2–CH2–COOH) and amino
caproic acid [NH 2 –(CH 2) 5 COOH] and is
biodegradable.
39. Which of the following species contains equal
number of s- and p- bonds :(1) XeO4
(2) (CN)2
(3) CH2(CN)2
(4) HCO3–
Ans. (1)
CODE-F
Ans. (3)
O
Sol.
Sol. H C
3
Xe
O
41.
Ans.
Sol.
43.
number of s bonds = 4
number of p bonds = 4
Which of these statements about [Co(CN)6]3– is true:(1) [Co(CN)6]3– has four unpaired electrons and will
be in a low-spin configuration.
(2) [Co(CN)6]3– has four unpaired electrons and will
be in a high spin configuration.
(3) [Co(CN)6]3– has no unpaired electrons and will
be in a high-spin configurtion.
(4) [Co(CN)6]3– has no unpaired electrons and will
be in a low-spin configuration.
(4)
[Co(CN)6]–3
Co+3 = 3d6 4s0 4p0
Q in presence of strong field ligand, pairing of
electrons occurs so in this complex no unpaired
electron is present and it is low spin complex.
Which one is not equal to zero for an ideal solution:(1) DSmix
(2) DVmix
(3) DP = Pobserved - PRaoult (4) DHmix
(1)
For an ideal solution DSmix > 0
Which property of colloidal solution is independent
of charge on the colloidal particles :(1) Electrophoresis
(2) Electro-osmosis
(3) Tyndall effect
(4) Coagulation
(3)
Tyndall effect is an optical property, so it is
independent of charge.
Given :-
OH
O
CH3
CH3
CH3
CH3
H
OH
O
H
CH3
CH3
Sol.
OH
45.
0.693
for first order t1/2 is independent of
K
concentration
A single compound of the structure :CH3 CH3
t1/2 =
OHC
C
C H C C O
H2
H2
is obtainable from ozonolysis of which of the following
cyclic compounds ?
(1)
CH3
CH3
O
(II)
O
(III)
CH3
CH3
Which of the given compounds can exhibit
tautomerism?
(1) I and III
(2) II and III
(3) I, II and III
(4) I and II
CH3
When initial concentration of a reactant is doubled
in a reaction, its half-life period is not affected. The
order of the reaction is :(1) First
(2) Second
(3) More than zero but less than first
(4) Zero
Ans. (1)
H3C
H3C
CH3 CH3
O
(I)
CH3
O
44.
A
Ans.
Sol.
42.
H
EN
Ans.
Sol.
CH3
O
LL
40.
O
H3C
CH3
(2)
H3C
CH3
(3)
CH3
CH3
H3C
(4)
CH3
Ans. (4)
Sol.
H3C
CH3
O3
Zn+H2 O
H3 C
O C
OHC
CH3
7
AIPMT-2015
46.
Which of the following endoparasites of humans
does show viviparity ?
(1) Enterobius vermicularis
(2) Trichinella spiralis
(3) Ascaris lumbricoides
(4) Ancylostoma duodenale
Ans. (2)
47. Cryopreservation of gametes of threatened species
in viable and fertile condition can be referred to as:(1) Advanced ex-situ conservation of biodiversity
(2) In situ conservation by sacred groves
(3) In situ cryo-conservation of biodiversity
(4) In situ conservation of biodiversity
Ans. (1)
48. Which one one of the following matches is correct ?
(1) Alternaria
Sexual
reproduction
absent
(2) Mucor
Reproduction Ascomycetes
by
Conjugation
(3) Agaricus
Parasitic
fungus
Ans.
52.
Ans.
53.
Basidiomycetes
Basidiomycetes
A
Ans. (1)
49. Minerals known to be required in large amounts for
plant growth include :(1) calcium, magnesium, manganese, copper
(2) potassium, phosphorus, selenium, boron
Ans.
54.
Ans.
55.
Ans.
56.
(3) magnesium, sulphur, iron, zinc
(4) phosphorus, potassium, sulphur, calcium
Ans. (4)
50.
Which of the following enhances or induces fusion
of protoplasts ?
Ans.
57.
(1) Polyethylene glycol and sodium nitrate
(2) IAA and kinetin
(3) IAA and gibberellins
(4) Sodium chloride and potassium chloride
Ans. (1)
8
Which of these is not an important component of
initiation of parturition in humans ?
(1) Synthesis of prostaglandins
(2) Release of oxytocin
(3) Release of prolactin
(4) Increase in estrogen and progesterone ratio
(3)
In which of the following gametophyte is not
independent free living ?
(2) Pteris
(1) Marchantia
(3) Pinus
(4) Funaria
(3)
Which of the following is not a sexually transmitted
disease ?
(1) Acquired Immuno Deficiency Syndrome (AIDS)
(2) Trichomoniasis
(3) Encephalitis
(4) Syphilis
(3)
Leaves become modified into spines in :(1) Pea
(2) Onion
(3) Silk Cotton
(4) Opuntia
(4)
Which one gives the most valid and recent
explanation for stomatal movements ?
(1) Potassium influx and efflux
(2) Starch hydrolysis
(3) Guard cell photosynthesis
(4) Transpiration
(1)
Which of the following had the smallest brain
capacity ?
(1) Homo sapiens
(2) Homo neanderthalensis
(3) Homo habilis
(4) Homo erectus
(3)
High value of BOD (Biochemical Oxygen Demand)
indicates that :(1) Water is highly polluted
(2) Water is less polluted
(3) Consumption of organic matter in the water is
higher by the microbes
(4) Water is pure
(1)
EN
Deuteromycetes
LL
(4) Phytophthora Aseptate
mycelium
51.
Ans.
CODE-F
58.
65.
Å
Ans.
60.
Ans.
61.
Ans.
62.
A
Ans.
63.
Ans.
64.
Ans.
67.
Ans.
68.
Ans.
69.
LL
Ans.
59.
66.
The crops engineered for glyphosate are resistant/
tolerant to :(1) Bacteria
(2) Insects
(3) Herbicides
(4) Fungi
(3)
Which of the following statements is not correct?
(1) Goblet cells are present in the mucosa of
intestine and secrete mucus
(2) Oxyntic cells are present in the mucosa of
stomach and secrete HCl.
(3) Acini are present in the pancreas and secrete
carboxypeptidase
(4) Brunner's glands are present in the submucosa
of stomach and secrete pepsinogen
(4)
In sea urchin DNA, which is double stranded, 17%
of the bases were shown to be cytosine. The
percentages of the other three bases expected to
be present in this DNA are :(1) G 17%, A 16.5%, T 32.5%
(2) G 17%, A 33%, T 33%
(3) G 8.5%, A 50%, T 24.5%
(4) G 34%, A 24.5%, T 24.5%
(2)
In Bt cotton, the Bt toxin present in plant tissue as
pro-toxin is converted into active toxin due to :(1) Acidic pH of the insect gut
(2) Action of gut micro-organisms
(3) Presence of conversion factors in insect gut
(4) Alkaline pH of the insect gut
(4)
Cytochromes are found in :(1) Outer wall of mitochondria
(2) Cristae of mitochondria
(3) Lysosomes
(4) Matrix of mitochondria
(2)
Read the following five statements (A to E) and select
the option with all correct statements :(A) Mosses and Lichens are the first organisms to
colonise a bare rock.
(B) Selaginella is a homosporous pteridophyte
(C) Coralloid roots in Cycas have VAM
(D) Main plant body in bryophytes is gametophytic,
whereas in pteridophytes it is sporophytic
(E) In gymnosperms, male and female gametophytes
are present within sporangia located on
sporophyte
(1) (B), (C) and (D)
(2) (A), (D) and (E)
(3) (B), (C) and (E)
(4) (A), (C) and (D)
(2)
9
EN
Ans.
Sliding filament theory can be best explained as :(1) Actin and Myosin filaments shorten and slide pass
each other
(2) Actin and Myosin filaments do not shorten but
rather slide pass each other
(3) When myofilaments slide pass each other, Myosin
filaments shorten while Actin filaments do not
shorten
(4) When myofilaments slide pass each other Actin
filaments shorten while Myosin filament do not
shorten
(2)
A gymnast is able to balance his body upside down
even in the total darkness because of :(1) Vestibular apparatus (2) Tectorial membrane
(3) Organ of corti
(4) Cochlea
(1)
A man with blood group 'A' marries a woman with
blood group 'B'. What are all the possible blood
groups of their offsprings ?
(1) A,B and AB only
(2) A,B,AB and O
(3) O only
(4) A and B only
(2)
Typical growth curve in plants is :(1) Linear
(2) Stair-steps shaped
(3) Parabolic
(4) Sigmoid
(4)
The UN Conference of Parties on climate change
in the year 2011 was held in :(1) South Africa
(2) Peru
(3) Qatar
(4) Poland
(1)
A technique of micropropagation is :(1) Somatic embryogenesis
(2) Protoplast fusion
(3) Embryo rescue
(4) Somatic hybridization
(1)
How many pairs of contrasting characters in pea
plants were studied by Mendel in his experiments ?
(1) Six
(2) Eight
(3) Seven
(4) Five
(3)
Ans.
70.
Ans.
71.
K(5) C(5) A5 G(2) is the floral formula of :-
(1) Sesbania
(3) Brassica
(2) Petunia
(4) Allium
Ans. (2)
Ans.
AIPMT-2015
Which one of the following is correct ?
(1) Serum = Blood + Fibrinogen
(2) Lymph = Plasma + RBC + WBC
(3) Blood = Plasma + RBC + WBC
(4) Plasma = Blood – Lymphocytes
Ans. (3)
73. The movement of a gene from one linkage group
to another is called :(1) Duplication
(2) Translocation
(3) Crossing over
(4) Inversion
Ans. (2)
74. Which body of the Government of India regulates
GM research and safety of introducing GM
organisms for public services ?
(1) Indian Council of Agricultural Research
(2) Genetic Engineering Approval Committee
(3) Research Committee on Genetic Manipulation
(4) Bio-safety committee
Ans. (2)
75. Rachel Carson's famous book "Silent Spring" is
related to :(1) Noise pollution
(2) Population explosion
(3) Ecosystem management
Ans.
80.
Ans.
81.
Ans.
82.
Ans.
83.
Ans.
84.
LL
(4) Pesticide pollution
Ans. (4)
76. Gastric juice of infants contains :-
79.
(1) nuclease, pepsinogen, lipase
(2) pepsinogen, lipase, rennin
(3) amylase, rennin, pepsinogen
A
(4) maltase, pepsinogen, rennin
Ans. (2)
77. Which of the following is not one of the prime health
risks associated with greater UV radiation through
the atmosphere due to depletion of stratospheric
ozone ?
(1) Reduced Immune System
(2) Damage to eyes
(3) Increased liver cancer
Ans.
85.
Ans.
86.
(4) Increased skin cancer
Ans. (3)
78. Capacitation refers to changes in the :(1) Ovum before fertilization
(2) Ovum after fertilization
(3) Sperm after fertilization
(4) Sperm before fertilization
Ans. (4)
10
Most animals are tree dwellers in a:(1) Thorn woodland
(2) Temperate deciduous forest
(3) Tropical rain forest
(4) Coniferous forest
(3)
True nucleus is absent in :(1) Mucor
(2) Vaucheria
(3) Volvox
(4) Anabaena
(4)
Glenoid cavity articulates :(1) Scapula with acromion
(2) Clavicle with scapula
(3) Humerus with scapula
(4) Clavicle with acromion
(3)
Transmission tissue is characteristic feature of :(1) Solid style
(2) Dry stigma
(3) Wet stigma
(4) Hollow style
(1)
DNA is not present in :(1) Ribosomes
(2) Nucleus
(3) Mitochondria
(4) Chloroplast
(1)
Gene regulation governing lactose operon of E.coli
that involves the lac I gene product is :
(1) Negative and inducible because repressor
protein prevents transcription
(2) Negative and repressible because repressor
protein prevents transcription
(3) Feedback inhibition because excess of
b-galactosidase can switch off trascription
(4) Positive and inducible because it can be induced
by lactose
(1)
Which of the following does not favour the formation
of large quantities of dilute urine ?
(1) Caffeine
(2) Renin
(3) Atrial-natriuretic factor (4) Alcohol
(2)
What causes a green plant exposed to the light on
only one side, to bend toward the source of light as
it grows ?
(1) Green plants seek light because they are
phototropic
(2) Light stimulates plant cells on the lighted side to
grow faster
(3) Auxin accumulates on the shaded side,
stimulating greater cell elongation there.
(4) Green plants need light to perform photosynthesis
(3)
EN
72.
Ans.
CODE-F
87.
93.
Nuclear envelope is a derivative of :(1) Membrane of Golgi complex
(2) Microtubules
(3) Rough endoplasmic reticulum
(4) Smooth endoplasmic reticulum
Ans. (3)
88. Select the correct option :I
(1) Leucocytes
(2) Helper T- Lymphocytes
(3) Thrombocytes
(4) B- Lymphocytes
Ans. (2)
94. Hysteresctomy is surgical removal of :
II
(a)
Synapsis aligns
homologous chromosomes
(i)
Anaphase-II
(b)
Synthesis of RNA and
protein
(ii)
Zygotene
(c)
Action of enzyme
recombinase
(iii) G2-phase
(d)
Centromeres do not
separate but chromatids
move towards opposite
poles
(iv)
Ans.
91.
Ans.
92.
Ans.
Pachytene
(a)
(b)
(c)
(d)
(1) (ii)
(iii)
(v)
(iv)
(2) (i)
(ii)
(v)
(iv)
(3) (ii)
(iii)
(iv)
(v)
(4) (ii)
(i)
(iii)
(iv)
(1)
Keel is the characteristic feature of flower of :(1) Indigofera (2) Aloe (3) Tomato (4) Tulip
(1)
Perigynous flowers are found in :(1) Cucumber
(2) China rose
(3) Rose
(4) Guava
(3)
A chemical signal that has both endocrine and
neural roles is ?
(1) Calcitonin
(2) Epinephrine
(3) Cortisol
(4) Melatonin
(2)
In which of the following both pairs have correct
combination :(1) In situ conservation : Cryopreservation
Ex situ conservation : Wildlife Sanctuary
(2) In situ conservation : Seed Bank
Ex situ conservation : National Park
(3) In situ conservation : Tissue culture
Ex situ conservation : Sacred groves
(4) In situ conservation : National Park
Ex situ conservation : Botanical Garden
(4)
A
Ans.
90.
(4) Uterus
Ans. (4)
95. Removal of proximal convoluted tubule from the
nephron will result in:
(1) More concentrated urine
(2) No change in quality and quantity of urine
(3) No urine formation
(4) More diluted urine
Ans. (4)
96. A major characteristic of the monocot root is the
presence of :
(1) Scattered vascular bundles
LL
Ans.
89.
Anaphase-I
(1) Prostate gland
(2) Vas-deference
(3) Mammary glands
EN
(v)
HIV that causes AIDS, first starts destroying:
(2) Vasculature without cambium
(3) Cambium sandwiched between phloem and
xylem along the radius
(4) Open vascular bundles
Ans. (2)
97.
Which of the following characteristics is mainly
responsible for diversification of insects on land ?
(1) Bilateral symmetry
(2) Exoskeleton
(3) Eyes
(4) Segmentation
Ans. (2)
98. Which of the following cells during gametogenesis
is normally dipoid?
(1) Spermatid
(2) Spermatogonia
(3) Secondary polar body
(4) Primary polar body
Ans. (2)
99. The structures that are formed by stacking of
organized flattened membranous sacs in the
chloroplasts are :
(1) Grana
(2) Stroma lamellae
(3) Stroma
(4) Cristae
Ans. (1)
11
AIPMT-2015
107. Erythropoiesis starts in :
(1) Liver
(2) Spleen
(3) Red bone marrow
(4) Kidney
Ans. (3)
108. Which is the most common mechanism of genetic
variation in the population of sexually reproducing
organism?
(1) Chromosomal aberrations
(2) Genetic drift
(3) Recombination
(4) Transduction
Ans. (3)
109. Blood pressure in the mammalian aorta is
maximum during :
EN
100. The chromosomes in which centromere is situated
close to one end are:
(1) Acrocentric
(2) Telocentric
(3) Sub-metacentric
(4) Metacentric
Ans. (1)
101. In a ring girdled plant:
(1) The root dies first
(2) The shoot and root die together
(3) Neither root nor shoot will die
(4) The shoot dies first
Ans. (1)
102. Vertical distribution of different species occupying
different levels in a biotic community is known as:
(1) Stratification
(2) Zonation
(3) Pyramid
(4) Divergence
Ans. (1)
103. Multiple alleles are present :
(1) At different loci on the same chromosome
(2) At the same locus of the chromosome
(3) On non-sister chromatids
(4) On different chromosomes
Ans. (2)
104. The mass of living material at a trophic level at a
particular time is called :
(1) Standing state
(2) Net primary productivity
(3) Standing crop
(4) Gross primary productivity
Ans. (3)
(1) Diastole of the right ventricle
(2) Systole of the left ventricle
(3) Diastole of the right atrium
(4) Systole of the left atrium
Ans. (2)
A
LL
110. When you hold your breath, which of the following
gas changes in blood would first lead to the urge to
breathe?
105. Which of the following animals is not viviparous?
(1) Elephant
(2) Platypus
(3) Whale
(4) Flying fox (Bat)
Ans. (2)
106. In an ecosystem the rate of production of organic
matter during photosynthesis is termed as:
(1) Gross primary productivity
(2) Secondary productivity
(3) Net productivity
(4) Net primary productivity
Ans. (1)
12
(1) rising CO2 concentration
(2) falling CO2 concentration
(3) rising CO2 and falling O2 concentration
(4) falling O2 concentration
Ans. (1)
111. Vascular bundles in monocotyledons are considered
closed because:
(1) Cambium is absent
(2) There are no vessels with perforations
(3) Xylem is surrounded all around by phloem
(4) A bundle sheath surrounds each bundle
Ans. (1)
112. Male gemetes are flagellated in :
(1) Anabaena
(2) Ectocarpus
(3) Spirogyra
(4) Polysiphonia
Ans. (2)
CODE-F
113. Which one of the following may require pollinators,
but is genetically similar to autogamy ?
(1) Xenogamy
(2) Apogamy
(3) Cleistogamy
(4) Geitonogamy
Ans. (4)
114. In ginger vegetative propagation occurs through:
(1) Offsets
(2) Bulbils
(3) Runners
(4) Rhizome
Ans. (4)
115. Which one of the following is not an inclusion body
120. Which of the following represents the correct
combination without any exception?
Characteristics
(1)
Mouth ventral, gills without
operculum; skin with placoid
scales; persistent notochord
Chondrichthyes
(2)
Sucking and circular mouth; jaws
absent, integument without scales;
paired appendages
Cyclostomata
(3)
Body covered with feathers; skin
moist and glandular; fore-limbs
form wings; lungs with air sacs
Aves
(4)
Mammary gland; hair on body;
pinnae; two pairs of Limbs
Mammalia
found in prokaryotes ?
(1) Cyanophycean granule
(2) Glycogen granule
(3) Polysome
(4) Phosphate granule
Ans. (3)
Ans. (1)
121. Which one of the following statements in incorrect?
(1) In competitive inhibition, the inhibitor molecule
is not chemically changed by the enzyme.
EN
116. A somatic cell that has just completed the S phase
(2) The competitive inhibitor does not affect the rate
of breakdown of the enzyme-substrate
complex.
of its cell cycle, as compared to gamete of the same
species, has :
(1) same number of chromosomes but twice the
amount of DNA
(3) The presence of the competitive inhibitor
decreases the Km of the enzyme for the
substrate.
(2) twice the number of chromosomes and four times
the amount of DNA
LL
(3) four times the number of chromosomes and twice
the amount of DNA
(4) twice the number of chromosomes and twice the
amount of DNA
Ans. (2)
117. Alleles are :
(1) true breeding homozygotes
(2) different molecular forms of a gene
(3) heterozygotes
(4) different phenotype
Ans. (2)
118. Select the correct matching in the following pairs:
(1) Smooth ER – Synthesis of lipids
(2) Rough ER– Synthesis of glycogen
(3) Rough ER – Oxidation of fatty acids
(4) Smooth ER – Oxidation of phospholipids
Ans. (1)
119. The terga, sterna and pleura of cockroach body are
joined by :
(1) Muscular tissue
(2) Arthrodial membrane
(3) Cartilage
(4) Cementing glue
Ans. (2)
A
Class
(4) A competitive inhibitor reacts reversibly withthe
enzyme to form an enzyme- inhibitor complex.
Ans. (3)
122. Which of the following regions of the brain is
incorrectly paired with its function?
(1) Cerebellum- language comprehension
(2) Corpus callosum-communication between the left
and right cerebral cortices
(3) Cerebrum- calculation and contemplation
(4) Medulla oblongata - homeostatic control
Ans. (1)
123. Which one of the following statements is not true?
(1) Pollen grains of some plants cause severe
allergies and bronchial afflictions in some people
(2) The flowers pollinated by flies and bats secrete
foul odour to attract them
(3) Honey is made by bees by digesting - pollen
collected from flowers
(4) Pollen grains are rich in nutrients, and they are
used in the form of tablets and syrups
Ans. (3)
13
AIPMT-2015
131. The following graph depicts changes in two
populations (A and B) of herbivores in a grassy field
A possible reason for these changes is that:
(4) erythrocytes; mucosa and submucosa of colon
(4) Fucus spp.
A
Time
(1) Population B competed more successfully for
food than population A
(2) Population A produced more offspring than
population B
(3) Population A consumed the members of
population B
(4) Both plant populations in this habitat decreased
Ans. (1)
132. Match each disease with its correct type of vaccine:
EN
Ans. (4)
125. Which of the following viruses is not transferred
through semen of an infected male?
(1) Human immunodeficiency virus
(2) Chikungunya virus
(3) Ebola virus
(4) Hepatitis B virus
Ans. (2)
126. A population will not exist in Hardy - Weinberg
equilibrium if :
(1) There are no mutations
(2) There is no migration
(3) The population is large
(4) Individuals mate selectively
Ans. (4)
127. The guts of cow and buffalo possess:
(1) Chlorella spp.
(2) Methanogens
(3) Cyanobacteria
A
LL
Ans. (2)
128. The hilum is a scar on the :
(1) Fruit, where it was attached to pedicel
(2) Fruit, where style was present
(3) Seed, where micropyle was present
(4) Seed, where funicle was attached
Ans. (4)
129. Secondary Succession takes place on/in :
(1) Degraded forest
(2) Newly created pond
(3) Newly cooled lava
(4) Bare rock
Ans. (1)
130. Which one of the following statements is wrong?
(1) Agar - agar is obtained from Gelidium and
Gracilaria
(2) Chlorella and Spirulina are used as space food
(3) Mannitol is stored food in Rhodophyceae
(4) Algin and carragen are products of algae
Ans. (3)
14
B
Number of
Organisms
124. The active from of Entamoeba histolytica feeds
upon:
(1) mucosa and submucosa of colon only
(2) food in intestine
(3) blood only
(a) tuberculosis
(i) harmless virus
(b) whooping cough
(ii) inactivated toxin
(c) diphtheria
(iii) killed bacteria
(d) polio
(iv) harmless bacteria
(a)
(b)
(c)
(d)
(1) (iii)
(2) (iv)
(ii)
(iii)
(iv)
(ii)
(i)
(i)
(3) (i)
(ii)
(iv)
(iii)
(4) (ii)
(i)
(iii)
(iv)
Ans. (2)
133. Which of the following are the important floral
rewards to the animal pollinators?
(1) Nectar and pollen grains
(2) Floral fragrance and calcium crystals
(3) Protein pellicle and stigmatic exudates
(4) Colour and large size flower
Ans. (1)
134. An abnormal human baby with 'XXX' sex
chromosomes was born due to :
(1) formation of abnormal ova in the mother
(2) fusion of two ova and one sperm
(3) fusion of two sperms and one ovum
(4) formation of abnormal sperms in the father
Ans. (1)
135. Transpiration and root pressure cause water to rise
in plants by :
(1) Pulling and pushing it, respectively
(2) Pushing it upward
(3) Pushing and pulling it, respectively
(4) Pushing it upward
Ans. (1)
CODE-F
136. An electron moving in a circular orbit of radius r
makes n rotations per second. The magnetic field
produced at the centre has magnitude :
(1) Zero
m 0n 2e
r
(2)
(3)
m0ne
2r
(4)
138. When tw o dispalcements represen ted by
y1 = asin(wt) and y2 = b cos(wt) are superimposed
the motion is :
m0ne
2pr
(1) simple harmonic with amplitude
Ans. (3)
Sol. Magnetic field due to a circular loop
B=
m0 NI
2r
(2) simple harmonic with amplitude
(3) simple harmonic with amplitude
Where N ® no. of loops
I=
q
e
=
= ne
T 1/n
B=
m0 ne
(Here N = 1 as e– makes only one loop)
2r
& y2 = b cos wt = b sin (wt +
EN
B
Amplitude A =
so A =
LL
P(in kPa) 2
4
6
3
V(in m )
A
The change in internal energy of the gas during the
transition is :
(1) –20 kJ (2) 20 J
(3) –12 kJ (4) 20 kJ
Ans. (1)
& T=
Sol. DU = n CV DT
so DT = T2 - T1 =
so
PV
nR
P2 V2 - P1 V1
nR
nR æ P2 V2 - P1 V1 ö P2 V2 - P1 V1
DU =
÷=
g - 1 çè
g -1
nR
ø
Þ DU =
-8 ´ 10
2/5
Hence option (1)
=–20 kJ
p
)
2
A12 + A 22 + 2A1 A 2 cos f
p
2
a2 + b2
Hence option (2)
139. A particle of unit mass undergoes one-dimensional
motion such that its velocity varies according to
v(x) = bx–2n
where b and n are constants and x is the position
of the particle. The acceleraion of the particle as
a function of x, is given by :
(1) –2nb2x–4n–1
(2) –2b2x–2n+1
(3) –2nb2e–4n+1
(4) –2nb2x–2n–1
Ans. (1)
Sol. v = b x–2n
so
dv
= – 2nb x–2n–1
dx
Now a = v
3
(a + b)
2
since the frequencies for both SHM are same,
resultant motion will be SHM. Now
here A1 = a, A2 = b & f =
A
a2 + b2
(4) not a simple harmonic
Ans. (2)
Sol. y1 = a sin wt
hence option (3)
137. One mole of an ideal diatomic gas undergoes a
transition from A to B along a path AB as shown
in the figure,
5
a
b
dv
= (bx–2n) (–2nb x–2n–1)
dx
Þ a = –2nb2 x–4n–1
Hence option (1)
15
AIPMT-2015
140. If radius of the
125
53
the radius of
(1)
27
13
Al nucleus is taken to be RAl then
= 2D
Te nucleus is nearly :
5
R Al
3
(2)
1/3
æ 13 ö
(3) ç
÷
è 53 ø
Sol. Linear width of central maxima = D (2q) = 2Dq
l
a
3
R Al
5
1/3
æ 53 ö
(4) ç
÷
è 13 ø
R Al
R Al
q
q
Ans. (1)
Sol. R µ A1/3
D
1/ 3
R Al æ 27 ö
=ç
÷
R Te è 125 ø
Þ RTee =
5
R
3 Al
Hence Option (4)
143. Across a metallic conductor of non-uniform cross
section a constant potential difference is applied.
The quantity which remains constant along the
conductor is :
(1) current
(2) drift velocity
(3) electric field
(4) current density
Ans. (1)
Sol. Metallic conductor can be considered as the
combination of various conductors connected in
series combination. And in series combination the
current always remains constant.
LL
EN
hence option (1)
141. In a double slit experiment, the two slits are 1 mm
apart and the screen is placed 1 m away. A
monochromatic light of wavelength 500 nm is used.
What will be the width of each slit for obtaining ten
maxima of double slit within the central maxima of
single slit pattern ?
(1) 0.1 mm
(2) 0.5 mm
(3) 0.02 mm
(4) 0.2 mm
Ans. (4)
Sol. Angular width of central maxima in double slit
lD
b
l
experiment =
= d =
D
D
d
A
Angular width of central maxima in single slit
experiment =
2l
d'
According to the question
10l 2l
=
d
d'
Þ d' = 0.2 d = 0.2 mm
hence option (4)
142. For a parallel beam of monochromatic light of
wavelength 'l', diffraction is produced by a single
slit whose width 'a' is of the order of the wavelength
of the light. If 'D' is the distance of the screen from
the slit, the width of the central maxima will be :
(1)
Ans. (4)
16
Dl
a
(2)
Da
l
(3)
2Da
l
(4)
2Dl
a
Hence option (1)
144. On observing light from three different stars P, Q
and R, it was found that intensity of violet color is
maximum in the spectrum of P, the intensity of green
colour is maximum in the spectrum of R and the
intensity or red colour is maximum in the spectrum
of Q. If TP, TQ and TR are the respective absolute
temperatures of P, Q and R, then it can be
concluded from the above observation that :
(1) TP > TR > TQ
(2) TP < TR < TQ
(3) TP < TQ < TR
(4) TP > TQ > TR
Ans. (1)
Sol. From Wein's displacement law
1
lm µ
T
Now from sequence ‘VIBGYOR’
(lm)P < (lm)R < (lm)Q
So TP > TR > TQ
Hence option (1)
CODE-F
145. A potentiometer wire has length 4 m and resistance
8W. The resistance that must be connected in series
with the wire and an accumulator of e.m.f. 2V, so
as to get a potential gradient 1 mV per cm on the
wire is :
(1) 40 W
(2) 44 W
(3) 48 W
147. A wire carrying current I has the shape as shown
in adjoining figure. Linear parts of the wire are very
long and parallel to X-axis while semicircular portion
of radius R is lying in Y-Z plane. Magnetic field at
point O is :
(4) 32 W
Z
Ans. (4)
Sol. Potential gradient =
1mV
= 10–3 V/cm = 10–1V/m
cm
I
R
I
2
I=
8+R
I
X
Potential drop across the potentiometer wire
Þ R = 32W
Hence option (4)
A
146. Consider 3 rd orbit of He + (Helium), using
non-relativistic approach, the speed of electron in
this orbit will be [given K = 9 × 109 constant,
Z = 2 and h (Planck's Constant) = 6.6 × 10–34 J s]
8
(3) 3.0 × 10 m/s
(4) 2.92 × 106 m/s
Ans. (1)
Sol. For H-like atoms
Z
× 2.188 × 106 m/s
n
here Z = 2, n = 3
6
n = 1.46 × 10 m/s
Hence option (1)
r
µ0 I ( ˆ
pi + 2kˆ )
4p R
r
µ0 I ( ˆ
p i - 2kˆ )
4p R
r
µ0 I ( ˆ
p i + 2kˆ )
4p R
(3) B =
LL
4
=
= 0.1
8+R
(2) 0.73 × 106 m/s
µ0 I ( ˆ
pi - 2kˆ )
4p R
(2) B = -
æ 16 ö 1
Potential gradient = ç
÷ ´ V/m
è8+Rø 4
(1) 1.46 × 106 m/s
r
(1) B = -
EN
8´2
16
=
=
8+R 8+R
n=
Y
O
Let the resistance to be connected is R then
(4) B =
Ans. (2)
Sol. 'B' due to segment '1'
B1 =
m0 I
[sin 90° + sin q] (– k̂ )
4 pR
2
m0 I
B1 =
(– k̂ ) = B3
4 pR
B due to segment '2'
B2 =
I 1
3
I
m0 I
( -ˆi)
4R
r
r
r
r
so 'B' at center Bc = B1 + B2 + B3
r
Þ BC =
-m 0 I æ
-m 0 I
2kˆ ö
ˆ
çè î + ÷ø =
( pˆi + 2k)
4R
4 pR
p
Hence option (2)
17
AIPMT-2015
148. Which of the following figures represent the
variation of particle momentum and the associated
de-Broglie wavelength ?
p
p
(1)
(2)
l
l
p
pipe =
Now
(4)
l
l
hc
1
Þ P µ (Rectangular hyperbola)
l
l
hence option (1)
(1) The energy stored in the capacitor decreases
K times.
A
1
æ1
ö
CV 2 ç – 1 ÷ .
2
èK
ø
(3) The charge on the capacitor is not conserved.
(4) The potential difference between the plates
decreases K times.
Ans. (3)
Sol. Once the capacitor is charged, its charge will be
constant Q = CV
When dielectric slab is inserted
C'New = KC
E=
1
Q2
Þ ENew =
E
K initial
2C
Q
1
V=
so Vnew =
V
C
K
Hence option (3)
18
v
3v
Þ l0 = 6lc = 6(20 cm) = 120 cm
=
4l c 2l 0
Hence option (2)
151. The refracting angle of a prism is A, and refractive
index of the material of the prism is cot(A/2). The
angle of minimum deviation is :(1) 180° – 2A
(2) 90° – A
(3) 180° + 2A
(4) 180° – 3A
Ans. (1)
æd +Aö
sin ç m
÷
è 2 ø
Sol. m =
æAö
sin ç ÷
è2ø
Q m = cot (A/2)
LL
149. A parallel plate air capacitor of capacitance C is
connected to a cell of emf V and then disconnected
from it. A dielectric slab of dielectric constant K,
which can just fill the air gap of the capacitor, is now
inserted in it. Which of the following is incorrect ?
(2) The change in energy stored is
3v
2l 0
EN
Ans. (1)
v
4l c
2nd overtone frequency of open organ pipe =
p
(3)
Sol. P =
150. The fundamental frequency of a closed organ pipe
of length 20 cm is equal to the second overtone of
an organ pipe open at both the ends. The length
of organ pipe open at both the ends is :(1) 100 cm
(2) 120 cm
(3) 140 cm
(4) 80 cm
Ans. (2)
Sol. Fun damental fre quen cy o f closed org an
æd +Aö
sin ç m
÷
è 2 ø
\ cot (A/2) =
æAö
sin ç ÷
è2ø
d +Aö
Þ cos (A/2) = sin æç m
÷
è 2 ø
dm + A
2
Þ dm = 180° – 2A
hence option (1)
152. Which logic gate is represented by the following
combination of logic gates ?
Þ 90° – A/2 =
A
Y1
Y
B
(1) NAND
(3) NOR
Ans. (2)
Y2
(2) AND
(4) OR
CODE-F
Sol. y1 = A , y2 = B ,
y = y 1 + y 2 = A + B (using De-morgan's theorem)
y=A·B
Hence this logic gate represents AND gate.
Hence option (2)
1
as
153. A Carnot engine, having an efficiency of h =
10
heat engine, is used as a refrigerator. If the work
done on the system is 10 J, the amount of energy
absorbed from the reservoir at lower temperature
is :(1) 99 J
(2) 90 J
(3) 1 J
(4) 100 J
Ans. (2)
Sol. For Engine & refrigerators operating between two
same temperatures
1
1+ b
Þ
(1)
E/C
E/C
2E
C2
Sol.
l
6
(4) 6 l
Ans. (1)
hc hc
Sol. eVS = E – f Þ VS = le – l e
0
here
hc hc
3V0 = le – l e
0
hc
hc
and V0 = 2le – l e
0
equation (1) – 3 × equation (2)
hc 2hc
+
Þ 0=–
2le l 0e
Þ l0 = 4l
Hence option (1)
(4)
E
C
E
C
So momentum transferred to the surface
2E
= pƒ – pi =
C
Hence option (1)
Momentum of light p =
attached to a string which passes through a smooth
hole in the plane as shown.
LL
A
(3)
E
C2
plane with velocity v0 at a radius R0. The mass is
Hence option (2)
154. A certain metallic surface is illuminated with
monochromatic light of wavelength, l. The stopping
potential for photo-electric current for this light is
3V0. if the same surface is illuminated with light of
wavelength 2l, the stopping potential is V0. The
threshold wavelength for this surface for photoelectric
effect is :l
4
(3)
Ans. (1)
Q2
(From the principle of refrigerator)
W
Q
9= 2
Þ Q2 = 90 Joule
10
(2)
(2)
156. A mass m moves in a circle on a smooth horizontal
1
1
=
Þb=9
10 1 + b
b=
(1) 4 l
2E
C
EN
h=
155. A radiation of energy ‘E’ falls normally on a perfectly
reflecting surface. The momentum transferred to
the surface is (C = Velocity of light) :-
...(1)
...(2)
v0
m
R0
The tension in the string is increased gradually and
R
finally m moves in a circle of radius 0 . The final
2
value of the kinetic energy is :(1)
1
mv 20
4
(2) 2mv 20
(3)
1
mv 20
2
(4) mv 20
Ans. (2)
Sol. Angular momentum remains Constant because of
the torque of tension is zero.
Þ Li = Lf
Þ mv0R = mv
R
2
Þ v = 2v0
KEf =
1
2
m ( 2v0 ) = 2mv 02
2
hence option (2)
19
AIPMT-2015
157. Two identical thin plano-convex glass lenses
(refractive index 1.5) each having radius of
curvature of 20 cm are placed with their convex
surfaces in contact at the centre. The intervening
space is filled with oil of refractive index 1.7. The
focal length of the combination is :(1) –25 cm
(2) –50 cm
(3) 50 cm
(4) –20 cm
Ans. (2)
a
Sol.
m1
A
mk
T
mk m1g
T
m2
B
a
m2g
n2 = 1.7
For the motion of both blocks
m2g – T = m2a
T – mk m1g = m1a
Sol.
n1 = 1.5
Þ a=
For the block of mass 'm2'
From lens maker's formula
é m - mk m1 ù
m2 g - T = m2 ê 2
úg
ë m1 + m2 û
EN
æ 1
1
1 ö
= (m - 1) ç
÷
f
R
R
è 1
2 ø
we have
é m - m k m1 ù
é m1 + mk m1 ù
T = m2 g - ê 2
ú m2 g = m2 g ê
ú
ë m1 + m2 û
ë m1 + m2 û
1
æ 1 ö 1
= (1.5 - 1) ç
÷=
f1
è 20 ø 40
1
æ 1 ö 1
= (1.5 - 1) ç
÷=
f2
è 20 ø 40
Now
1
7
æ 2 ö
= (1.7 - 1) ç
=
÷
f3
è 20 ø 100
1 1 1 1
= + +
f f1 f2 f3
1 1
1
7
=
+
f 40 40 100
Þ
f = – 50 cm
Hence option (2)
158. A block A of mass m1 rests on a horizontal table.
A
Þ
Þ T=
LL
&
(m2 - mk m1 )g
m1 + m2
A light string connected to it passes over a
frictionless pulley at the edge of table and from its
other end another block B of mass m2 is suspended.
The coefficient of kinetic friction between the block
and the table is µ k. When the block A is sliding on
the table, the tension in the string is :-
m1m2 (1 + mk )g
m 1 + m2
Hence option (2)
159. A particle is executing SHM along a straight line.
Its velocities at distances x1 and x2 from the mean
position are V1 and V2, respectively. Its time period
is :x 22 – x12
(1) 2p
V12 – V22
V12 + V22
(2) 2p
x12 + x 22
V12 – V22
(3) 2p
x12 – x 22
x12 + x 22
(4) 2p
V12 + V22
Ans. (1)
Sol. For particle undergoing SHM
V = w A 2 - x2
so V1 = w A 2 - x12
solving these two equations we get
V12 - V22 2p
=
x 22 - x12
T
(1)
(m2 – mk m1 )g
(m1 + m2 )
(2)
m1m2 (1 + m k )g
(m1 + m2 )
w=
(3)
m1m2 (1 – m k )g
(m1 + m2 )
(4)
(m2 + mk m1 )g
(m1 + m2 )
Þ T = 2p
Ans. (2)
20
& V2 = w A 2 - x22
x 22 - x12
V12 - V22
Hence option (1)
CODE-F
160. A ship A is moving Westwards with a speed of
10 km h–1 and a ship B 100 km South of A, is
moving Northwards with a speed of 10 km h–1. The
time after which the distance between them
becomes shortest, is :(1) 5 h
(2) 5 2 h
(3) 10 2 h
(4) 0 h
Ans. (1)
Sol. As we know
10 kmph
B=
A
ph
vB
A
=
10
2
Now P = rgh & compressibility 'K' =
so
r
v BA = 102 + 102 = 10 2 kmph
A
tsn = 5 hrs.
Hence option (1)
161. A rod of weight W is supported by two parallel knife
edges A and B and is in equilibrium in a horizontal
position. The knives are at a distance d from each
other. The centre of mass of the rod is at distance
x from A. The normal reaction on A is :-
(3)
(2)
W(d – x)
d
(4)
Ans. (3)
Sol. By torque balancing
about B
NA(d) = W(d–x)
Þ NA =
Hence option (2)
163. Two particles of masses m1, m2 move with initial
velociteis u1 and u2. On collision, one of the particles
get excited to higher level, after abosrbing energy
e. If final velocities of particles be v1 and v2 then we
must have :
LL
50 2 50 2
=
r
v BA
10 2
Wd
x
W(d - x)
d
Hence option (3)
DV
= rgh (K)
V
= 1.201 × 10–2
Time taken to reach the shortest distance between
(1)
1
B
= 103 × 9.8 × 2700 × 45.4 × 10–11
distance OB = 100 cos45°= 50 2 km
A&B=
DV P
=
V
B
km
45°
B
so
O
10 kmph
P
DV
V
EN
Sol.
100km
162. The approximate depth of an ocean is 2700 m. The
compressibility of water is 45.4 × 10–11 Pa–1 and
density of water is 103 kg/m3.What fractional
compression of water will be obtained at the bottom
of the ocean ?
(1) 1.0 × 10–2
(2) 1.2 × 10–2
–2
(3) 1.4 × 10
(4) 0.8 × 10–2
Ans. (2)
(1)
1
1
1
1
m 1u12 + m 2 u 22 = m 1v12 + m 2 v 22 - e
2
2
2
2
(2)
1
1
1
1
m 1u12 + m 2 u 22 - e = m 1v12 + m 2v 22
2
2
2
2
(3)
1 2 2 1 2 2
1
1
m 1 u1 + m 2 u 2 + e = m 12v12 + m 22 v 22
2
2
2
2
W(d – x)
x
Wx
d
(4) m12u1 + m22u2 – e = m12v1 + m22v2
Ans. (2)
NA
NB
Sol. Energy will always be conserved so
K. E.initial = K.E.final + Excitation energy
A
B
x
W
d–x
1
1
1
1
m1u12 + m2u22 = m1v12 + m2 v 22 + e
2
2
2
2
Hence option (2) .
21
AIPMT-2015
164. Kepler's third law states that square of period of
revolution (T) of a planet around the sun, is
proportional to third power of average distance r
between sun and planet
i.e. T2 = Kr3
here K is constant.
If the masses of sun and planet are M and m
respectively then as per Newton's law of gravitation
force of attraction between them is
F =
GMm
, here G is gravitational constant.
r2
From work energy theorem W = Kƒ – Ki
Þ –25 = Kƒ –
Þ Kƒ = 475
Hence option (4)
166. A wind with speed 40 m/s blows parallel to the roof
of a house. The area of the roof is 250 m2.
Assuming that the pressure inside the house is
atmospheric pressure, the force exerted by the wind
on the roof and the direction of the force will be :
(rair = 1.2 kg/m3)
The relation between G and K is described as :
(1) GMK = 4p2
(2) K = G
(4) GK = 4p2
Ans. (2)
æ
ç as n =
è
EN
2pr
2pr
r
=
Sol. T =
v
GM
Sol. By Bernaulli's equation
1
P + rv2 = P0 + 0
2
GM ö
÷
r ø
2p
GM
1
p0 - p = rv2
2
3/2
r
2
LL
4p
· r3
GM
Comparing
4p 2
GM
Hence option (1)
165. A block of mass 10 kg, moving in x direction with
a constant speed of 10 ms–1, is subjected to a
retarding force F = 0.1 x J/m during its travel from
x = 20 m to 30 m. Its final KE will be :
A
K=
(3) 2.4 × 105 N, downwards
(4) 4.8 × 105 N, downwards
Ans. (1)
T2 =
(1) 4.8 × 105 N, upwards
(2) 2.4 × 105 N, upwards
1
(3) K =
G
T=
1
2
10(10)
2
(1) 450 J
(2) 275 J
(3) 250 J
(4) 475 J
Ans. (4)
Sol. W = – ò Fdx
Pin = P0
vin = 0
1
F = rv2 A
2
F = 2.4 × 105 upward
Hence Option (2)
167. Two spherical bodies of mass M and 5 M and radii
R and 2 R are released in free space with initial
separation between their centres equal to 12 R. If
they attract each other due to gravitational force
only, then the distance covered by the smaller body
before collision is :
(1) 4.5 R
(2) 7.5 R
(3) 1.5 R
(4) 2.5 R
Ans. (2)
12R
Sol.
30
W = – ò 0.1x dx
Initial distance between their centers = 12 R
20
30
é x2 ù
W = –0.1 ê ú
ë 2 û 20
é 900 – 400 ù
W = –0.1 ê
úû = –25
ë
2
22
R
2R
At time of collision the distance between their
centers = 3R
So total distance travelled by both=12R–3R = 9R
CODE-F
Since the bodies move under mutual forces, center
of mass will remain stationary so
CP
= g in terms of
CV
169. The ratio of the specific heats
m1x1 = m2x2
degrees of freedom (n) is given by :
mx = 5m(9R –x)
x = 45R – 5x
6x = 45R
(1) ç 1 +
æ
è
nö
÷
3ø
(2) ç 1 +
æ
è
2ö
÷
nø
45
R
6
(3) ç 1 +
æ
è
nö
÷
2ø
(4) ç 1 +
æ
è
1ö
÷
nø
x=
Ans. (2)
x = 7.5R
(1) P
R
Z
æRö
÷
èZø
(2) P ç
æRö
÷
èZø
(4) P ç
(3) P
Ans. (4)
Sol.
R
V2
R
Þ V2 = PR
2
n
2
P
4
6×10 Pa
Resistor & Inductor
circuit
R
L
f
Impedance = Z
éVù
P ' = V. ê ú .cos f
ë Zû
P' =
V2 R
·
Z Z
(From phasor diagram)
P' =
(PR)R
Z2
2
æRö
P' = ç ÷ P
è Zø
Hence option (4)
B
C
(Phasor diagram)
A
P=
\ g= 1 +
Hence option (2)
170. Figure below shows two paths that may be taken
by a gas to go from a state A to a state C.
R
Z
AC source
2
n
Here degree of freedom ® n
LL
Pure resistor
circuit
Sol. g = 1 +
EN
Hence option (2)
168. A resitance 'R' draws power 'P' when connected to
an AC source. If an inductance is now placed in
series with the resistance, such that the impedance
of the circuit becomes 'Z', the power drawn will be:
XL
A
2×104 Pa
–3
3
2×10 m
4×10–3 m3
V
In process AB, 400 J of heat is added to the system
and in process BC, 100 J of heat is added to the
system. The heat absorbed by the system in the
process AC will be :
(1) 500 J
(2) 460 J
(3) 300 J
(4) 380 J
Ans. (2)
Sol. In cyclic process ABCA,
DUcyclic = 0
Qcyclic = Wcyclic
QAB + QBC + QCA = closed loop area.
400 + 100 + QCA =
1
× (2 × 10–3) × 4 × 104
2
400 + 100 –QAC = 40
QAC = 460 J
Hence option (2)
23
AIPMT-2015
171. If energy (E), velocity (V) and time (T) are chosen as
the fundamental quantities, the dimensional formula
of surface tension will be :
(1) [EV–1T–2]
(2) [EV–2T–2]
(3) [E–2V–1T–3]
(4) [EV–2T–1]
Ans. (2)
Sol. Applying dimensional analysis
a
173. Three blocks A, B and C of masses 4 kg, 2 kg and
1 kg respectively, are in contact on a frictionless
surface, as shown. If a force of 14 N is applied on
the 4 kg block, then the contact force between A
and B is :
b c
SµEVT
1 0 –2
1 2 –2 a
1 0 –2
a 2a –2a
1 –1 b
1c
M L T = k [M L T ] [L T ] [T ]
M L T = k [M L T
Comparision
a =1
(1) 6 N
Ans. (1)
b –b c
]L T T
2a + b = 0
–2 = –2a – b + c
b = –2
–2 = –2(1) + 2 + c
(2) 8 N
(3) 18 N
Sol. Acceleration of system
c = –2
So the dimensional formula for surface tension will
1 –2 –2
be [E V T ]
Alternate solution :
Surface energy
Surface Tension =
Area
[E]
–2 –2
[Surface tension] =
= [E V T ]
[V × T]2
Hence option (2)
172. If in a p-n junction, a square input signal of 10 V
is applied as shown,
Fnet
= M
total
=
EN
14 N
4kg
(4) 2 N
14
= 2 m/s2
4 + 2 +1
2kg 1kg
The contact force between 4 kg & 2 kg block will
move 2 kg & 1 kg block with the same acceleration
so Fcontact = (2 + 1)a = 3(2) = 6N
+5V
LL
Hence option (1)
174. A, B and C are voltmeters of resistance R, 1.5 R
RL
–5V
A
then the output across RL will be :
and 3R respectively as shown in the figure. When
some potential difference is applied between X and
Y, the voltmeter readings are V A, VB and VC
respectively. Then :
10V
(1)
(2)
B
–5V
5V
(3)
(4)
–10V
Ans. (3)
Sol. This is the circuit where P-N junction is acting as a
Half-wave rectifier so the output will be
5V
Hence option (3)
24
X
A
(1) VA ¹ VB = VC
(3) VA ¹ VB ¹ VC
Ans. (4)
Y
C
(2) VA = VB ¹ VC
(4) VA = VB = VC
Sol. Effective resistance of B & C =
(1.5R)(3R)
=R
1.5R + 3R
In series sequence V µ R
so voltage across 'A' = voltage across B & C
Now B & C are parallel so VB = VC
Þ VA = VB = VC
Hence option (4)
CODE-F
175. Three identical spherical shells, each of mass m and
radius r are placed as shown in figure. Consider an
axis XX' which is touching to two shells and passing
through diameter of third shell.
Moment of inertia of the system consisting of these
three spherical shells about XX' axis is :X
177. The two ends of a metal rod are matainted at
temperatures 100°C and 110°C. The rate of heat
flow in the rod is found to be 4.0 J/s. If the ends
are maintained at temperatures 200°C and 210°C,
the rate of heat flow will be :
(1) 16.8 J/s
(2) 8.0 J/s
(3) 4.0 J/s
(4) 44.0 J/s
Ans. (3)
Sol. Rate of heat flow µ temperature difference
between two ends
X'
(1) 3 mr2
(2)
16
mr 2
5
(3) 4 mr2
(4)
11 2
mr
5
Þ
Ans. (3)
X
dQ
µ (T2 - T1 )
dt
Here temperature differnece in both case is same
(i.e. 10°C)
1
Sol.
2
Ixx'
r
r
EN
So, rate of heat flow will also be same
3
So,
X'
= I1 + I2 + I3
Hence option (3)
178. Two similar springs P and Q have spring constants
KP and KQ, such that KP > KQ. The are stretched,
first by the same amount (case a,) then by the same
force (case b). The work done by the springs WP and
W Q are related as, in case (a) and case (b),
respectively :
(1) WP = WQ ; WP = WQ
(2) WP > WQ ; WQ > WP
(3) WP < WQ ; WQ < WP
(4) WP = WQ ; WP > WQ
Ans. (2)
Sol. Given KP > KQ
Case (a) : x1 = x2 = x
LL
2 2 æ2 2
ö æ2
ö
mr + ç mr + mr 2 ÷ + ç mr 2 + mr 2 ÷
3
è3
ø è3
ø
dQ
= 4 J/s
dt
A
(Using parallel axis theorem)
Þ Ixx' = 2mr2 + 2mr2 = 4mr2
Hence option (3)
176. The electric field in a certain region is acting radially
outward and is given by E = Ar. A charge contained
in a shepere of radius 'a' centred at the origin of
the field, will be given by :
(1) A e0 a2
(2) 4 pe0 Aa3
3
(3) e0 Aa
(4) 4 pe0 Aa2
Ans. (2)
r
Sol. Flux linked with sphere = E.dsr
since electric field is radial. It is always perpendicular
to the surface.
so f = Ar.(4 pr 2 )
f = A(a)(4pr 2 )
f = A4pa
(as r = a)
3
Now according to gauss law
f=
q in
Î0
Þ
qin = f.Î0
1
K x2
WP 2 P
K
=
= P Þ WP > WQ
WQ 1
K
Q
K Q x2
2
Case (b) : F1 = F2 = F
For constant force
W=
F2
1
ÞWµ
2K
K
so q in = A4pa Î0
WP K Q
So W = K Þ WQ > WP
Q
P
Hence option (2)
Hence option (2)
3
25
AIPMT-2015
179. A conducting square frame of side 'a' and a long
straight wire carrying current I are located in the
same plane as shown in the figure. The frame moves
to the right with a constant velocity 'V'. The emf
induced in the frame will be proportional to :
180. A particle of mass m is driven by a machine that
delivers a constant power k watts. If the particle
starts from rest the force on the particle at time t
is :
(1)
x
(3)
I
V
mk t
-1
2
1
-1
mk t 2
2
-1
(2)
2mk t
(4)
mk -1 2
t
2
Ans. (4)
Sol. P = Fv = mav
dv
dt
By integrating the equation
Þ K = mv
a
1
(2)
(2x + a )2
1
(3)
(2x - a )(2x + a )
(4)
1
x2
Ans. (3)
2
dv
2k æ 1 - 12 ö
=
t ÷
dt
m çè 2
ø
æ 1 ö 2k
F = ma = m ç ÷
è 2 ø mt
a
V
x + a/2
A
emf Induced in side (1)
e1 = B1Vl
emf Induced is side (2)
e2 = B2Vl
emf in the frame = B1Vl – B2Vl
e = Vl [B1 – B2]
Þ e µ B1 - B 2
Since B µ
é
ù
ê 1
1 ú
So e µ ê
a
aú
êx x+ ú
êë
2
2 úû
é
ù
1
1
ú
ëê ( 2x - a ) ( 2x + a ) ûú
Þ eµê
Hence Option (3)
26
2k
v2
k
t
= t Þ v=
m
2 m
a=
LL
x–a/2 1
Sol.
Þ
Hence option (4)
x
I
ò v dv = ò m dt
EN
1
(1)
(2x - a )2
k
Þ
1
r
Þ F=
mk
2t
2